Question

What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.

Name: What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.
Uploaded: 2022-06-27T10:44:25-08:00
Duration: 4 min 36 s
Channel: Adi S
Description: What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.

          What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.

What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.

Added by -Ngeles H.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

06/27/2022

Step 1

Given that Ka = 1.8 * 10^(-5) for CH3COOH, we can use the equation for Ka to find the concentration of H+ ions: Ka = [H+][CH3COO-] / [CH3COOH] 1.8 * 10^(-5) = x * 0.26 / 0.29 x = (1.8 * 10^(-5) * 0.29) / 0.26 x = 2 * 10^(-5) M Show more…

Show all steps

Thanks for your feedback!

What is the final pH if 0.71 mol NaOH is added to a buffer solution that is 1.00 L of a 0.29 M CH3COOH and 0.26 M CH3COO-? Ka of CH3COOH is 1.8 x 10^-5. Assume the volume of the solution does not change with the addition of NaOH.